Water Is Leaking Out Of An Inverted Conical Tank At A Rate Of $11,500 \, \text{cm}^3/\text{min}$ At The Same Time That Water Is Being Pumped Into The Tank At A Constant Rate. The Tank Has A Height Of 6 M And The Diameter At The Top Is 4 M.

Introduction

In this article, we will delve into a mathematical problem involving an inverted conical tank. The tank is leaking water at a constant rate, while water is being pumped into the tank at the same time. Our goal is to determine the rate at which water is being pumped into the tank to counteract the leakage.

Problem Statement

The tank has a height of 6 m and a diameter of 4 m at the top. Water is leaking out of the tank at a rate of $11,500 , \text{cm}^3/\text{min}$. We need to find the rate at which water is being pumped into the tank to maintain a constant water level.

Mathematical Formulation

Let's denote the radius of the water surface at time $t$ as $r(t)$ and the height of the water surface as $h(t)$. Since the tank is conical, the volume of water in the tank at time $t$ is given by:

$$V(t) = \frac{1}{3} \pi r^2 h$$

We know that the tank is leaking water at a rate of $11,500 , \text{cm}^3/\text{min}$. This means that the rate of change of the volume of water in the tank is equal to the negative of the leakage rate:

$$\frac{dV}{dt} = -11,500 \, \text{cm}^3/\text{min}$$

We also know that water is being pumped into the tank at a constant rate. Let's denote this rate as $k$. Then, the rate of change of the volume of water in the tank is also equal to the pumping rate:

$$\frac{dV}{dt} = k$$

Equations of Motion

Since the tank is conical, the radius of the water surface at time $t$ is related to the height of the water surface by:

$$r(t) = \frac{2}{3} h(t)$$

Substituting this into the equation for the volume of water in the tank, we get:

$$V(t) = \frac{1}{3} \pi \left(\frac{2}{3} h(t)\right)^2 h(t) = \frac{2}{27} \pi h^3(t)$$

Differentiating this with respect to time, we get:

$$\frac{dV}{dt} = \frac{2}{9} \pi h^2(t) \frac{dh}{dt}$$

Substituting this into the equation for the rate of change of the volume of water in the tank, we get:

$$\frac{2}{9} \pi h^2(t) \frac{dh}{dt} = -11,500 \, \text{cm}^3/\text{min}$$

Solving the Equations

We can solve this equation for $\frac{dh}{dt}$:

$$\frac{dh}{dt} = -\frac{11,500 \, \text{cm}^3/\text{min}}{\frac{2}{9} \pi h^2(t)}$$

We also know that the height of the water surface is related to the radius of the water surface by:

$$h(t) = \frac{3}{2} r(t)$$

Substituting this into the equation for $\frac{dh}{dt}$, we get:

$$\frac{dh}{dt} = -\frac{11,500 \, \text{cm}^3/\text{min}}{\frac{2}{9} \pi \left(\frac{3}{2} r(t)\right)^2}$$

Simplifying this, we get:

$$\frac{dh}{dt} = -\frac{11,500 \, \text{cm}^3/\text{min}}{\frac{27}{4} \pi r^2(t)}$$

Finding the Pumping Rate

We know that the rate of change of the volume of water in the tank is equal to the pumping rate:

$$\frac{dV}{dt} = k$$

Substituting this into the equation for $\frac{dV}{dt}$, we get:

$$k = \frac{2}{9} \pi h^2(t) \frac{dh}{dt}$$

Substituting the expression for $\frac{dh}{dt}$, we get:

$$k = \frac{2}{9} \pi h^2(t) \left(-\frac{11,500 \, \text{cm}^3/\text{min}}{\frac{27}{4} \pi r^2(t)}\right)$$

Simplifying this, we get:

$$k = -\frac{4000}{27} \pi h^2(t) \frac{1}{r^2(t)}$$

Numerical Solution

To find the pumping rate, we need to know the initial height and radius of the water surface. Let's assume that the initial height is 6 m and the initial radius is 2 m.

Substituting these values into the equation for $k$, we get:

$$k = -\frac{4000}{27} \pi (6 \, \text{m})^2 \frac{1}{(2 \, \text{m})^2}$$

Simplifying this, we get:

$$k = -\frac{4000}{27} \pi (36 \, \text{m}^2) \frac{1}{4 \, \text{m}^2}$$

$$k = -\frac{4000}{27} \pi (9 \, \text{m}^2)$$

$$k = -\frac{36000}{27} \pi \, \text{m}^2/\text{min}$$

$$k = -\frac{4000}{3} \pi \, \text{m}^2/\text{min}$$

Conclusion

In this article, we have analyzed a mathematical problem involving an inverted conical tank. We have found the rate at which water is being pumped into the tank to counteract the leakage. The pumping rate is given by:

$$k = -\frac{4000}{3} \pi \, \text{m}^2/\text{min}$$

Q: What is the problem with the inverted conical tank?

A: The tank is leaking water at a constant rate, while water is being pumped into the tank at the same time. We need to find the rate at which water is being pumped into the tank to maintain a constant water level.

Q: What are the dimensions of the tank?

A: The tank has a height of 6 m and a diameter of 4 m at the top.

Q: What is the rate at which water is leaking out of the tank?

A: The tank is leaking water at a rate of $11,500 , \text{cm}^3/\text{min}$.

Q: How do we find the rate at which water is being pumped into the tank?

A: We use the equation for the rate of change of the volume of water in the tank, which is given by:

$$\frac{dV}{dt} = k$$

where $k$ is the pumping rate.

Q: How do we relate the rate of change of the volume of water in the tank to the height and radius of the water surface?

A: We use the equation for the volume of water in the tank, which is given by:

$$V(t) = \frac{1}{3} \pi r^2 h$$

Differentiating this with respect to time, we get:

$$\frac{dV}{dt} = \frac{2}{9} \pi h^2(t) \frac{dh}{dt}$$

Q: How do we find the rate at which the height of the water surface is changing?

A: We use the equation for the rate of change of the height of the water surface, which is given by:

$$\frac{dh}{dt} = -\frac{11,500 \, \text{cm}^3/\text{min}}{\frac{2}{9} \pi h^2(t)}$$

Q: How do we find the pumping rate?

A: We substitute the expression for $\frac{dh}{dt}$ into the equation for the rate of change of the volume of water in the tank:

$$k = \frac{2}{9} \pi h^2(t) \frac{dh}{dt}$$

Simplifying this, we get:

$$k = -\frac{4000}{27} \pi h^2(t) \frac{1}{r^2(t)}$$

Q: What is the numerical solution for the pumping rate?

A: To find the pumping rate, we need to know the initial height and radius of the water surface. Let's assume that the initial height is 6 m and the initial radius is 2 m.

Substituting these values into the equation for $k$, we get:

$$k = -\frac{4000}{27} \pi (6 \, \text{m})^2 \frac{1}{(2 \, \text{m})^2}$$

Simplifying this, we get:

$$k = -\frac{4000}{27} \pi (36 \, \text{m}^2) \frac{1}{4 \, \text{m}^2}$$

$$k = -\frac{4000}{27} \pi (9 \, \text{m}^2)$$

$$k = -\frac{36000}{27} \pi \, \text{m}^2/\text{min}$$

$$k = -\frac{4000}{3} \pi \, \text{m}^2/\text{min}$$

Q: What is the conclusion of this problem?

A: In this article, we have analyzed a mathematical problem involving an inverted conical tank. We have found the rate at which water is being pumped into the tank to counteract the leakage. The pumping rate is given by:

$$k = -\frac{4000}{3} \pi \, \text{m}^2/\text{min}$$

This result can be used to design a system to pump water into the tank at the correct rate to maintain a constant water level.